To add two or more fractions that have the same denominators, add the numerators and place the resulting sum over the common denominator. Reduce, if necessary.
Find the following sums.
3737 size 12{ { {3} over {7} } } {} +
2727 size 12{ { {2} over {7} } } {}
The denominators are the same.
Add the numerators and place the sum over the common denominator, 7.
3737 size 12{ { {3} over {7} } } {} +
2727 size 12{ { {2} over {7} } } {} =
3+273+27 size 12{ { {3+2} over {7} } } {} =
5757 size 12{ { {5} over {7} } } {}
When necessary, reduce the result.
1818 size 12{ { {1} over {8} } } {} +
3838 size 12{ { {3} over {8} } } {} =
1+381+38 size 12{ { {1+3} over {8} } } {} =
4848 size 12{ { {4} over {8} } } {} =
1212 size 12{ { {1} over {2} } } {}
We do not add denominators.
To see what happens if we mistakenly add the denominators as well as the numerators, let’s add
1212 size 12{ { {1} over {2} } } {} and
1212 size 12{ { {1} over {2} } } {}.
Adding the numerators and mistakenly adding the denominators produces:
1212 size 12{ { {1} over {2} } } {} +
1212 size 12{ { {1} over {2} } } {} =
1+12+21+12+2 size 12{ { {1+1} over {2+2} } } {} =
2424 size 12{ { {2} over {4} } } {} =
1212 size 12{ { {1} over {2} } } {}
This means that
1212 size 12{ { {1} over {2} } } {} +
1212 size 12{ { {1} over {2} } } {}is the same as
1212 size 12{ { {1} over {2} } } {} , which is preposterous! We do not add denominators.
3838 size 12{ { {3} over {8} } } {} +
3838 size 12{ { {3} over {8} } } {}
6868 size 12{ { {6} over {8} } } {} =
3434 size 12{ { {3} over {4} } } {}
711711 size 12{ { {7} over {"11"} } } {} +
411411 size 12{ { {4} over {"11"} } } {}
11111111 size 12{ { {"11"} over {"11"} } } {} = 1
15201520 size 12{ { {"15"} over {"20"} } } {} +
120120 size 12{ { {1} over {"20"} } } {} +
220220 size 12{ { {2} over {"20"} } } {}
18201820 size 12{ { {"18"} over {"20"} } } {} =
910910 size 12{ { {9} over {"10"} } } {}
To subtract two or more fractions that have the same denominators, subtract the numerators and place the resulting difference over the common denominator. Reduce, if necessary.
Find the following differences.
3535 size 12{ { {3} over {5} } } {} -
1515 size 12{ { {1} over {5} } } {}
The denominators are the same.
Subtract the numerators and place the difference over the common denominator, 5.
3535 size 12{ { {3} over {5} } } {} -
1515 size 12{ { {1} over {5} } } {} =
3−153−15 size 12{ { {3 - 1} over {5} } } {} =
2525 size 12{ { {2} over {5} } } {}
When necessary, reduce the result.
8686 size 12{ { {8} over {6} } } {} -
2626 size 12{ { {2} over {6} } } {} =
6666 size 12{ { {6} over {6} } } {} = 1
We do not subtract denominators.
To see what happens if we mistakenly subtract the denominators as well as the numerators, let’s subtract
715715 size 12{ { {7} over {"15"} } } {} -
415415 size 12{ { {4} over {"15"} } } {}.
Subtracting the numerators and mistakenly subtracting the denominators produces:
715715 size 12{ { {7} over {"15"} } } {} -
415415 size 12{ { {4} over {"15"} } } {} =
7−415−157−415−15 size 12{ { {7 - 4} over {"15" - "15"} } } {} =
3030 size 12{ { {3} over {0} } } {}
We end up dividing by zero, which is undefined. We do not subtract denominators.
512512 size 12{ { {5} over {"12"} } } {} -
112112 size 12{ { {1} over {"12"} } } {}
412412 size 12{ { {4} over {"12"} } } {} =
1313 size 12{ { {1} over {3} } } {}
316316 size 12{ { {3} over {"16"} } } {} -
316316 size 12{ { {3} over {"16"} } } {}
165165 size 12{ { {"16"} over {5} } } {} -
1515 size 12{ { {1} over {5} } } {} -
2525 size 12{ { {2} over {5} } } {}
Result is
135135 size 12{ { {"13"} over {5} } } {}
Basic Rule: Fractions can only be added or subtracted conveniently if they have like denominators.
To see why this rule makes sense, let’s consider the problem of adding a quarter and a dime.
A quarter is
1414 size 12{ { {1} over {4} } } {} of a dollar.
A dime is
110110 size 12{ { {1} over {"10"} } } {} of a dollar.
We know that 1 quarter + 1 dime = 35 cents. How do we get to this answer by adding
1414 size 12{ { {1} over {4} } } {} and
110110 size 12{ { {1} over {"10"} } } {} ?
We convert them to quantities of the same denomination.
A quarter is equivalent to 25 cents, or
2510025100 size 12{ { {"25"} over {"100"} } } {}.
A dime is equivalent to 10 cents, or
1010010100 size 12{ { {"10"} over {"100"} } } {}.
By converting them to quantities of the same denomination, we can add them easily:
2510025100 size 12{ { {"25"} over {"100"} } } {} +
1010010100 size 12{ { {"10"} over {"100"} } } {} =
3510035100 size 12{ { {"35"} over {"100"} } } {}.
Same denomination
→
→
size 12{ rightarrow } {}
same denominator
If the denominators are not the same, make them the same by building up the fractions so that they both have a common denominator. A common denominator can always be found by multiplying all the denominators, but it is not necessarily the Least Common Denominator.
The LCD is the smallest number that is evenly divisible by all the denominators.
It is the least common multiple of the denominators.
The LCD is the product of all the prime factors of all the denominators, each factor taken the greatest number of times that it appears in any single denominator.
Find the sum of these unlike fractions.
112112 size 12{ { {1} over {"12"} } } {} +
415415 size 12{ { {4} over {"15"} } } {}
Factor the denominators:
12 = 2 × 2 × 3
15 = 3 × 5
What is the greatest number of times the prime factor 2 appear in any single denominator? Answer: 2times. That is the number of times the prime factor 2 will appear as a factor in the LCD.
What is the greatest number of times the prime factor 3 appear in any single denominator? Answer: 1 time. That is the number of times the prime factor 3 will appear as a factor in the LCD.
What is the greatest number of times the prime factor 5 appear in any single denominator? Answer: 1 time. That is the number of times the prime factor 5 will appear as a factor in the LCD.
So we assemble the LCD by multiplying each prime factor by the number of times it appears in a single denominator, or:
2 × 2 × 3 × 5 = 60
60 is the Least Common Denominator (the Least Common Multiple of 12 and 15).
To create fractions with like denominators, we now multiply the numerators by whatever factors are missing when we compare the original denominator to the new LCD.
In the fraction
112112 size 12{ { {1} over {"12"} } } {}, we multiply the denominator 12 by 5 to get the LCD of 60. Therefore we multiply the numerator 1 by the same factor (5).
112112 size 12{ { {1} over {"12"} } } {} ×
5555 size 12{ { {5} over {5} } } {} =
560560 size 12{ { {5} over {"60"} } } {}
Similarly,
415415 size 12{ { {4} over {"15"} } } {} ×
4444 size 12{ { {4} over {4} } } {} =
16601660 size 12{ { {"16"} over {"60"} } } {}
We can now add the two fractions because they have like denominators:
560560 size 12{ { {5} over {"60"} } } {} +
16601660 size 12{ { {"16"} over {"60"} } } {} =
21602160 size 12{ { {"21"} over {"60"} } } {}
Reduce the result:
21602160 size 12{ { {"21"} over {"60"} } } {} =
720720 size 12{ { {7} over {"20"} } } {}
1616 size 12{ { {1} over {6} } } {} +
3434 size 12{ { {3} over {4} } } {}
Result is
11121112 size 12{ { {"11"} over {"12"} } } {}
5959 size 12{ { {5} over {9} } } {} -
512512 size 12{ { {5} over {"12"} } } {}
Result is
536536 size 12{ { {5} over {"36"} } } {}
15161516 size 12{ { {"15"} over {"16"} } } {} +
1212 size 12{ { {1} over {2} } } {} -
3434 size 12{ { {3} over {4} } } {}
Result is
35163516 size 12{ { {"35"} over {"16"} } } {}
915915 size 12{ { {9} over {"15"} } } {} +
415415 size 12{ { {4} over {"15"} } } {}
Result is
13151315 size 12{ { {"13"} over {"15"} } } {}
710710 size 12{ { {7} over {"10"} } } {} -
310310 size 12{ { {3} over {"10"} } } {} +
11101110 size 12{ { {"11"} over {"10"} } } {}
Result is
15101510 size 12{ { {"15"} over {"10"} } } {} (reduce to 1
1212 size 12{ { {1} over {2} } } {} )
Find the total length of the screw in this diagram:
Total length is
19321932 size 12{ { {"19"} over {"32"} } } {} in.
5252 size 12{ { {5} over {2} } } {} +
162162 size 12{ { {"16"} over {2} } } {} -
3232 size 12{ { {3} over {2} } } {}
Result is
182182 size 12{ { {"18"} over {2} } } {} (reduce to 9)
3434 size 12{ { {3} over {4} } } {} +
1313 size 12{ { {1} over {3} } } {}
Result is
13121312 size 12{ { {"13"} over {"12"} } } {}
Two months ago, a woman paid off
324324 size 12{ { {3} over {"24"} } } {} of a loan. One month ago, she paid off
424424 size 12{ { {4} over {"24"} } } {} of the loan. This month she will pay off
524524 size 12{ { {5} over {"24"} } } {} of the total loan. At the end of this month, how much of her total loan will she have paid off?
She will have paid off
12241224 size 12{ { {"12"} over {"24"} } } {}, or
1212 size 12{ { {1} over {2} } } {} of the total loan.
8383 size 12{ { {8} over {3} } } {} -
1414 size 12{ { {1} over {4} } } {} +
736736 size 12{ { {7} over {"36"} } } {}
Result is
94369436 size 12{ { {"94"} over {"36"} } } {} (reduce to
47184718 size 12{ { {"47"} over {"18"} } } {} )